blocks|key|240707|text|听起来你想要np.select从numpy|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|240708|import+numpy+as+np

conds+=+[df.score+>=4,+df.score+<=2]

choices+=+['happy',+'sad']

df['happiness']+=+np.select(conds,+choices,+default='ok')

>>>+df
++++name++score+happiness
0++Jason++++++1+++++++sad
1++Molly++++++3++++++++ok
2+++Tina++++++4+++++happy
3+++Jake++++++5+++++happy
4++++Amy++++++2+++++++sad|code-block|syntax|javascript|240709|注意:通过使用numpy+(或pd.np，取决于导入熊猫的方式)而不是仅使用np，可以避免显式导入np。|240710|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.select.html^0|6|9|G|5|6|9|0|0|0|7|5|F|5|12|2|1D|2|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Sounds like you want <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.select.html" rel="nofollow noreferrer"><code>np.select</code></a> from <code>numpy</code>

<pre><code>import numpy as np

conds = [df.score &gt;=4, df.score &lt;=2]

choices = ['happy', 'sad']

df['happiness'] = np.select(conds, choices, default='ok')

&gt;&gt;&gt; df
 name score happiness
0 Jason 1 sad
1 Molly 3 ok
2 Tina 4 happy
3 Jake 5 happy
4 Amy 2 sad
</code></pre>

Note: you can avoid explicitly importing <code>numpy</code> by using <code>pandas.np</code> (or <code>pd.np</code>, depending how you imported pandas) instead of just <code>np</code>

blocks|key|1700037|text|使用pd.cut|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1700038|pd.cut(df.score,[0,2,4,np.Inf],labels=['sad','ok','happy'])
Out[594]:+
0++++++sad
1+++++++ok
2+++++++ok
3++++happy
4++++++sad

#df['yourcol']=pd.cut(df.score,[0,2,4,np.Inf],labels=['sad','ok','happy'])|code-block|syntax|javascript|1700039|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/generated/pandas.cut.html^0|2|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Using <a href="https://pandas.pydata.org/pandas-docs/stable/generated/pandas.cut.html" rel="nofollow noreferrer">pd.cut</a>

<pre><code>pd.cut(df.score,[0,2,4,np.Inf],labels=['sad','ok','happy'])
Out[594]: 
0 sad
1 ok
2 ok
3 happy
4 sad

#df['yourcol']=pd.cut(df.score,[0,2,4,np.Inf],labels=['sad','ok','happy'])
</code></pre>

blocks|key|241403|text|问题是，apply将您的函数应用于列中的每个值。df不是are_you_ok内部的一个DataFrame，而是一个整数(在您的例子中)。自然，Python抱怨不能用['happiness']索引整数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|241404|不过，您的代码很容易修复。只需重写are_you_ok，使其与整数参数一起工作。|241405|In+[1]:+import+pandas+as+pd
In+[2]:+df+=+pd.DataFrame({'name':+['Jason',+'Molly',+'Tina',+'Jake',+'Amy'],+
+++...:+++++++++'score':+[1,+3,+4,+5,+2]})
+++...:+++++++++
In+[3]:+def+are_you_ok(x):
+++...:+++++if+x+>=+4:
+++...:+++++++++return+'happy'
+++...:+++++elif+x+<=+2:
+++...:+++++++++return+'sad'
+++...:+++++else:
+++...:+++++++++return+'ok'
+++...:+++++
In+[4]:+df['happiness']+=+df['score'].apply(are_you_ok)
In+[5]:+df
Out[5]:+
++++name++score+happiness
0++Jason++++++1+++++++sad
1++Molly++++++3++++++++ok
2+++Tina++++++4+++++happy
3+++Jake++++++5+++++happy
4++++Amy++++++2+++++++sad|code-block|syntax|javascript|241406|entityMap^0|4|5|O|2|S|A|17|9|2A|D|0|H|A|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|N|$]]

The problem is that <code>apply</code> applies your function to every single value in the column. <code>df</code> is not a <code>DataFrame</code> inside of <code>are_you_ok</code>, but (in your case) an integer. Naturally, Python is complaining that you cannot index into integers with <code>['happiness']</code>.

Your code is quite easy to fix, though. Just rewrite <code>are_you_ok</code> such that it works with integer arguments.

<pre><code>In [1]: import pandas as pd
In [2]: df = pd.DataFrame({'name': ['Jason', 'Molly', 'Tina', 'Jake', 'Amy'], 
 ...: 'score': [1, 3, 4, 5, 2]})
 ...: 
In [3]: def are_you_ok(x):
 ...: if x &gt;= 4:
 ...: return 'happy'
 ...: elif x &lt;= 2:
 ...: return 'sad'
 ...: else:
 ...: return 'ok'
 ...: 
In [4]: df['happiness'] = df['score'].apply(are_you_ok)
In [5]: df
Out[5]: 
 name score happiness
0 Jason 1 sad
1 Molly 3 ok
2 Tina 4 happy
3 Jake 5 happy
4 Amy 2 sad
</code></pre>

So I have a dataframe:

<pre><code>import pandas as pd

df = pd.DataFrame({'name': ['Jason', 'Molly', 'Tina', 'Jake', 'Amy'], 
 'score': [1, 3, 4, 5, 2]})
</code></pre>

And I want to create a new column based on the conditions in the 'score' column.

I tried it out like this

<pre><code>df['happiness'] = df['score']
def are_you_ok(df):
 if df['happiness'] &gt;= 4:
 return 'happy',
 elif df['happiness'] &lt;= 2:
 return 'sad',
 else:
 return 'ok'

df['happines'] = df['happiness'].apply(are_you_ok)
df
</code></pre>

When I try to run that though, all I get is:

<pre><code>TypeError: 'int' object is not subscriptable
</code></pre>

Can I not use this kind of function with an integer?

'int' object is not subscriptable after if statement

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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所以我有一个数据：import pandas as pddf = pd.DataFrame({'name': ['Jason', 'Molly', 'Tina', 'Jake', 'Amy'],         'score': [1, 3, 4, 5, 2]})我想根据“分数”栏中的条件创建一个新的列。我就这样试过了...

问‘'int’对象在if语句之后不可订阅
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问‘'int’对象在if语句之后不可订阅EN